Question about using linearity in proof of an corollary of Fatou’s lemma I am reading Erhan Çınlar's "Probability and Stochastics". The use of linearity (underlined in red) in the following corollary stuck me.

The Fatou's Lemma would give us $\mu(\liminf (f_n-g))\le\liminf\mu(f_n-g)$ (for notational simplicity, I omitted the indicator $1_A$), which is
$\mu((\liminf f_n)-g)\le\liminf\mu(f_n-g)$.
I guess the linearity is supposed to be applied to both side to turn the above inequality into
$\mu(\liminf f_n)-\mu g\le\liminf(\mu f_n-\mu g)=\liminf\mu f_n-\mu g$
so that $\mu g$ can be cancelled to get what we proposed to prove. However, the linearity in the textbook is as follows:

From the proposition, there are two cases in which we can apply linearity correctly: 1) when two integrands are both in $\mathcal{E}_+$ and the coefficient is nonnegative. 2) when two integrands are both integrable. Looking at the left side first, since $g$ is integrable, 2) seems to be the case to apply, but in that case, $\liminf f_n$ must also be integrable. However, I could not get to this from the conditions in the corollary. If we are to use case 1), neither $\liminf f_n$ and $g$ are in $\mathcal{E}_+$, and what is worse, they are linearly combined by subtraction, not addition. So. I'm confused and don't know what kind of linearity in the proof the author means.
I tried two attempts to figure out what the linearity means.
1, I tried to prove an extended version of the linearity proposition to have $\mu((\liminf f_n)-g)=\mu (\liminf f_n)-\mu g$ using monotone convergence theorem, but I soon encountered a question: what on earth does $\mu(\liminf f_n)$ mean? The textbook defines three types of integrals:

Let's call it type a), b) and c), respectively. I can see no way to prove that $\liminf f_n\ge0$, so type b) is not applicable. I can't figure out either that $\liminf f_n$ satisfies type c). So, I'm stuck.
2, I tried to prove that $\liminf f_n$ is integrable using the domination similar to the proof of Lebesgue's dominated convergence theorem, but $f_n\ge g$ is not enough to be considered a domination that I can use to prove that $\liminf f_n$ is integrable. So I got stuck again.
I haven't yet talked about the right side of the equality, which is to show $\mu(f_n-g)=\mu f_n-\mu g$. I also have no idea how linearity can be applied to get this.
I hope anyone can help me figure out what the "linearity" mean in the proof of the corollary. I am self-studying so I don't have professor or TA to help me. It would be very good if you happen to have read this text or be using this book in a probability course. Please use the concept and notation in this book because I am just a new learner of measure-theoretical probability. Thank you.
 A: I solved the problem myself. There are two points that lead to the solution, as described below.

*

*Use domination and relation of negative part of functions to prove integral is well defined.

Proposition: If $f\ge g$ where $f\in\mathcal{E}$ and $g$ is integrable, then $f$ satisfies the integral of type c) (cited in the original question).
Prove: It can be proved that if two functions satisfy $f\ge g$, we have a reverse relation for their negative parts $f^-\le g^-$. Since $g$ is integrable, $\mu g^-$ is finite, so is $\mu f^-$ because $f^-\le g^-$. As a result, $\mu f$ is well-defined according to type c) of integral. $\blacksquare$
Since $f_n\ge g$ for all $n$, so $\liminf f_n\ge g$. Using the above proposition, we get $\liminf f_n$ satisfies type c) of integral. As a side note, $\liminf f_n\in\mathcal{E}$.


*Make a cautious extension of linearity.

Proposition 4.12 aims at type b) and integrable functions. It can be extended to type c) integral, provided that any part of the linear equation makes sense. Precisely,
Proposition: If $f$ and $g$ satisfy type c) of integral, we have, for any $a,b\in\mathbb{R}$, $\mu(af+bg)=a\mu f+b\mu g$, provided that any part of this equation is well-defined.
Proof is omitted because of many trivial cases. The purpose of the condition in the proposition is to avoid the undefined situation of $\infty-\infty$, which can happen if $af(x)=+\infty$ while $bg(x)=-\infty$ (not necessarily a.e.) on the left side or $a\mu f=+\infty$ while $b\mu g=-\infty$ on the right side. As long as these pathological situations do not occur, we can safely apply the linearity for type c) integral. A special case is when one of $f$ and $g$ is known to be integrable, as is in the Corollary.
Now we can safely and correctly write $\mu[\liminf(f_n-g)]=\mu[(\liminf f_n)-g]=\mu(\liminf f_n)-\mu g$, and $\liminf\mu(f_n-g)=\liminf(\mu f_n-\mu g)=(\liminf\mu f_n)-\mu g$, based on the above two points as well as the subadditivity of $\liminf$.
